The value of `lim_(n rarr infty) (1)/(n) {(n+1)(n+2)(n+3)…(n+n)}^(1//n)` is equal to

Options

A. `4e`
B. `(e)/(4)`
C. `(4)/(e)`
D. None of these

Answer

To solve the limit \( \lim_{n \to \infty} \frac{1}{n} \left( (n+1)(n+2)(n+3)\cdots(n+n) \right)^{\frac{1}{n}} \), we can follow these steps:

### Step 1: Rewrite the product
The product \( (n+1)(n+2)(n+3)\cdots(n+n) \) can be rewritten using the product notation:
\[ (n+1)(n+2)(n+3)\cdots(n+n) = \prod_{r=1}^{n} (n+r) \]

### Step 2: Factor out \( n \) from each term
We can factor \( n \) out of each term in the product:
\[ \prod_{r=1}^{n} (n+r) = \prod_{r=1}^{n} n\left(1 + \frac{r}{n}\right) = n^n \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \]

### Step 3: Substitute back into the limit
Now, substituting this back into our limit expression, we have:
\[ \lim_{n \to \infty} \frac{1}{n} \left( n^n \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}} = \lim_{n \to \infty} \frac{1}{n} \left(n^n\right)^{\frac{1}{n}} \left( \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}} \]
This simplifies to:
\[ \lim_{n \to \infty} \frac{1}{n} n \cdot \left( \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}} = \lim_{n \to \infty} \left( \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}} \]

### Step 4: Take the logarithm
Taking the logarithm of the product:
\[ \log y = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \log\left(1 + \frac{r}{n}\right) \]

### Step 5: Recognize the Riemann sum
As \( n \to \infty \), the sum becomes a Riemann sum for the integral:
\[ \log y = \int_{0}^{1} \log(1+x) \, dx \]

### Step 6: Evaluate the integral
To evaluate \( \int_{0}^{1} \log(1+x) \, dx \), we can use integration by parts:
Let \( u = \log(1+x) \) and \( dv = dx \).
Then \( du = \frac{1}{1+x} \, dx \) and \( v = x \).
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int_{0}^{1} \log(1+x) \, dx = \left[ x \log(1+x) \right]_{0}^{1} - \int_{0}^{1} \frac{x}{1+x} \, dx \]
Evaluating the first term:
\[ \left[ x \log(1+x) \right]_{0}^{1} = 1 \cdot \log(2) - 0 = \log(2) \]
Now we need to evaluate \( \int_{0}^{1} \frac{x}{1+x} \, dx \):
\[ \int_{0}^{1} \frac{x}{1+x} \, dx = \int_{0}^{1} \left(1 - \frac{1}{1+x}\right) \, dx = \left[ x - \log(1+x) \right]_{0}^{1} = 1 - \log(2) \]
Putting it all together:
\[ \int_{0}^{1} \log(1+x) \, dx = \log(2) - (1 - \log(2)) = 2\log(2) - 1 \]

### Step 7: Exponentiate to find \( y \)
Thus, we have:
\[ \log y = 2 \log 2 - 1 \implies y = e^{2 \log 2 - 1} = \frac{4}{e} \]

### Final Answer
The value of the limit is:
\[ \lim_{n \to \infty} \frac{1}{n} \left( (n+1)(n+2)(n+3)\cdots(n+n) \right)^{\frac{1}{n}} = \frac{4}{e} \]